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Question-187146

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Question Number 187146 by Rupesh123 last updated on 14/Feb/23
Answered by a.lgnaoui last updated on 14/Feb/23
△SMI et RNI semblables I centre de SRMN SK=((PQ)/2)=3 △BRQ RIO Semblables ((OR)/(OI))=(((RN)/2)/(HQ))=((RQ)/(BQ))⇒ ((RN)/(2HQ))=((RQ)/(BQ)) HQ=3 RQ=6 BQ=BP+6 ((RN)/6)=(6/(6+BP)) (1) △ASK ABH semblables cos tan 60=((AK)/(SK))=((AH)/(BH))⇒ ((AK)/3)=((AK+6)/(BP+3))=(√3) ⇒AK=3(√3) Alors ((3(√3) +6)/(BP+3))=(√3) BP=((3(√3) +6)/( (√3)))−3= BP=2(√3) (1)⇒RN=((36)/(6+BP))=((36)/(6+2(√3)))= RN=9−3(√3) Finalement:Aire totale(A) A=2×(Aire IRN) A=2(((RN×OI)/2))=RN×OI=27−9(√3) Aire=11,41
$$\bigtriangleup{SMI}\:\:{et}\:{RNI}\:{semblables}\:\:\: \\ $$$${I}\:{centre}\:{de}\:{SRMN}\:\:{SK}=\frac{{PQ}}{\mathrm{2}}=\mathrm{3} \\ $$$$\bigtriangleup{BRQ}\:\:\:{RIO}\:\:{Semblables} \\ $$$$\frac{{OR}}{{OI}}=\frac{\frac{{RN}}{\mathrm{2}}}{{HQ}}=\frac{{RQ}}{{BQ}}\Rightarrow\:\:\:\frac{{RN}}{\mathrm{2}{HQ}}=\frac{{RQ}}{{BQ}} \\ $$$${HQ}=\mathrm{3}\:\:{RQ}=\mathrm{6}\:\:\:{BQ}={BP}+\mathrm{6} \\ $$$$\frac{{RN}}{\mathrm{6}}=\frac{\mathrm{6}}{\mathrm{6}+{BP}}\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\bigtriangleup{ASK}\:\:\:\:{ABH}\:\:{semblables} \\ $$$$\mathrm{cos}\:\:\mathrm{tan}\:\:\mathrm{60}=\frac{{AK}}{{SK}}=\frac{{AH}}{{BH}}\Rightarrow \\ $$$$\frac{{AK}}{\mathrm{3}}=\frac{{AK}+\mathrm{6}}{{BP}+\mathrm{3}}=\sqrt{\mathrm{3}}\:\Rightarrow{AK}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${Alors}\:\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}\:+\mathrm{6}}{{BP}+\mathrm{3}}=\sqrt{\mathrm{3}} \\ $$$${BP}=\frac{\mathrm{3}\sqrt{\mathrm{3}}\:+\mathrm{6}}{\:\sqrt{\mathrm{3}}}−\mathrm{3}=\:\:\:\:\:\:{BP}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\left(\mathrm{1}\right)\Rightarrow{RN}=\frac{\mathrm{36}}{\mathrm{6}+{BP}}=\frac{\mathrm{36}}{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}}}= \\ $$$$\:\:\:\:\:\:\:\:{RN}=\mathrm{9}−\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${Finalement}:{Aire}\:{totale}\left({A}\right) \\ $$$${A}=\mathrm{2}×\left({Aire}\:{IRN}\right) \\ $$$${A}=\mathrm{2}\left(\frac{{RN}×{OI}}{\mathrm{2}}\right)={RN}×{OI}=\mathrm{27}−\mathrm{9}\sqrt{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:{Aire}=\mathrm{11},\mathrm{41} \\ $$$$ \\ $$
Commented by Rupesh123 last updated on 14/Feb/23
Excellent
Answered by mr W last updated on 14/Feb/23
say side length of equilateral is s. (((((√3)s)/2)−6)/(((√3)s)/2))=(6/s) ⇒s=6+4(√3) BP=(6/( (√3)))=2(√3) BQ=6+4(√3)−2(√3)=6+2(√3) ((TP)/(RQ))=((BP)/(BQ)) ⇒TP=((2(√3)×6)/(6+2(√3)))=3(√3)−3 ⇒ST=6−(3(√3)−3)=3(3−(√3)) A_(blue) =(6/2)×ST=9(3−(√3))≈11.412
$${say}\:{side}\:{length}\:{of}\:{equilateral}\:{is}\:{s}. \\ $$$$\frac{\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}−\mathrm{6}}{\frac{\sqrt{\mathrm{3}}{s}}{\mathrm{2}}}=\frac{\mathrm{6}}{{s}} \\ $$$$\Rightarrow{s}=\mathrm{6}+\mathrm{4}\sqrt{\mathrm{3}} \\ $$$${BP}=\frac{\mathrm{6}}{\:\sqrt{\mathrm{3}}}=\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${BQ}=\mathrm{6}+\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{\mathrm{3}}=\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\frac{{TP}}{{RQ}}=\frac{{BP}}{{BQ}} \\ $$$$\Rightarrow{TP}=\frac{\mathrm{2}\sqrt{\mathrm{3}}×\mathrm{6}}{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}}}=\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3} \\ $$$$\Rightarrow{ST}=\mathrm{6}−\left(\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\right)=\mathrm{3}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right) \\ $$$${A}_{{blue}} =\frac{\mathrm{6}}{\mathrm{2}}×{ST}=\mathrm{9}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)\approx\mathrm{11}.\mathrm{412} \\ $$
Commented by Rupesh123 last updated on 14/Feb/23
Excellent

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